I have a quantum formula describing what kind of photon should be emitted by an LED depending on its voltage. Of course the colour is depending on the material, but every type of LED also needs its specific voltage.
My formula uses 2.5V as an example and tells me that an LED working with 2.5V should emit photons with a wavelength of approximately 470nm, which is blue.
\begin{align}
\lambda &= \frac cf = \frac{ch}{E} = \frac{ch}{eU} \\
&= \frac{\rm 3\times10^8 \,\frac ms \cdot 6.626\times10^{-34}\,J\,s}
{\rm1.602\times10^{-19}\,C \cdot 2.5\,V}
\approx \rm 469\,nm
\end{align}
But in reality, blue LEDs need about 3.0V – 3.5V while 2.5V is enough for green LEDs!
Why does the equation not fit the reality and where goes my additional energy of about 0.5eV per photon? Is it converted to thermal energy (why and how?) or what happens with it?
Best Answer
What your formula is actually meant to give is the emission wavelength as a function of the material bandgap:
$$\lambda \approx \frac{\rm1240\,eV\,nm}{E_{gap}}.$$
In your example, when carriers recombine in a semiconductor with a bandgap of 2.5 eV, photons having a wavelength of 496 nm (blue-green) will be emitted.
The forward drop of an LED whose active-region material has a bandgap of 2.5 eV is usually significantly higher than 2.5 V. This means that the energy given to any electron injected in the active region is higher than the energy of the photon that you get in exchange. As a result you are heating up the device.
This happens when there is a significant series resistance. Only with ideal contacts, and p- and n-regions with zero resistivity, the forward drop of an LED can be equal to $E_{gap}/e$, so that no heating is caused.
By driving the LED with very small currents, when the voltage drop is smaller than $E_{gap}/e$, it is theoretically even possible to cool down the device. It is very difficult to put a real device in this condition: while the photon emission is cooling it down the small current is still heating it up, but it has been experimentally proven.